help with negative exponents

[Sdeb1129]

Problem: ( 8a to the negative 9th power ) multiplied to the negative 2/3 power

I multiplied -9 by - 2/3 and I got 18/3=6. So I believe that gives me a to the sixth power. Since it's a negative power the 8 becomes 1/8. But do I put it in a cube root with 8 to the -2 power? I don't know what to do with negative exponents in the denominator. Does it change to 8 to the 2/3 power and a to the 6th power?

 

[Mrspi]

Problem: ( 8a to the negative 9th power ) multiplied to the negative 2/3 power

I multiplied -9 by - 2/3 and I got 18/3=6. So I believe that gives me a to the sixth power. Since it's a negative power the 8 becomes 1/8. But do I put it in a cube root with 8 to the -2 power? I don't know what to do with negative exponents in the denominator. Does it change to 8 to the 2/3 power and a to the 6th power?

Is this your problem?

(8a[sup:2r77copk]-9[/sup:2r77copk])[sup:2r77copk](-2/3)[/sup:2r77copk]

If it is, then use the rule of exponents which says (ab)[sup:2r77copk]m[/sup:2r77copk] = a[sup:2r77copk]m[/sup:2r77copk] b[sup:2r77copk]m[/sup:2r77copk]

(8a[sup:2r77copk]-9[/sup:2r77copk])[sup:2r77copk](-2/3)[/sup:2r77copk] = 8[sup:2r77copk](-2/3)[/sup:2r77copk] (a[sup:2r77copk]-9[/sup:2r77copk])[sup:2r77copk](-2/3)[/sup:2r77copk]

And when you raise a power to a power, you multiply the exponents:

8[sup:2r77copk](-2/3)[/sup:2r77copk] a[sup:2r77copk]-9*(-2/3)[/sup:2r77copk]

or,

8[sup:2r77copk](-2/3)[/sup:2r77copk] a[sup:2r77copk]6[/sup:2r77copk]

Apply the definition of a negative exponent. a[sup:2r77copk]-n[/sup:2r77copk] = 1 / a[sup:2r77copk]n[/sup:2r77copk]

a[sup:2r77copk]6[/sup:2r77copk] / 8[sup:2r77copk](2/3)[/sup:2r77copk]

Evaluate 8[sup:2r77copk](2/3)[/sup:2r77copk] and you're finished.